Hint: it is sufficient to show that $U$ contains the commutator subgroup $[T, T]$ of $T$, i.e. that for any $X, Y \in T$, we have $XYX^{-1}Y^{-1} \in U$. This is not so hard: letting
$$X = \left( \begin{matrix} a & b \\ 0 & c \end{matrix} \right), Y = \left( \begin{matrix} d & e \\ 0 & f \end{matrix} \right)$$
you can actually compute $X^{-1}$ and $Y^{-1}$, and it is easy to see what happens to the diagonal entries.
I hate descriptions of a group like that in Herstein, because it is not at all obvious to a newcomer that there might not be some secret collapsing of everything into the identity element. How do you know you can just "declare" when two things are equal without creating some kind of inconsistency somewhere?
What Herstein is aiming to do looks like describing a group by generators and relations, but typically the most you can get from such a description alone is an upper bound on the size of the group, but to know the description is really a group of order $21$ you need to put in some work. To understand what I'm getting at, the post here gives a description of a group with 3 generators and a few relations that turns out to be the trivial group after some fairly nontrivial work.
Instead of trying to make sense of what Herstein is doing, let's instead just build ourselves a nonabelian group of order $21$ using $2 \times 2$ matrices with entries in $\mathbf Z/7\mathbf Z$.
In the group $(\mathbf Z/7\mathbf Z)^\times$, the subgroup of order $3$ is $\{1,2,4 \bmod 7\} = \{b \bmod 7 : b^3 \equiv 1 \bmod 7\}$.
Let $G = \{(\begin{smallmatrix}b&c\\0&1\end{smallmatrix}) \bmod 7 : b^3 = 1 \}$. The upper right entries of matrices in $G$ are arbitrary integers modulo $7$ while the upper left entries are restricted to be the elements of $(\mathbf Z/7\mathbf Z)^\times$ with order dividing $3$, and those are the powers of $2 \bmod 7$. As a set, $G$ has size $21$: $3$ options for the upper left entry and $7$ options for the upper right entry, and they can be picked arbitrarily.
Check $G$ is a closed under matix multiplication and inversion, so $G$ is a group of order $21$. In $G$, let $a = (\begin{smallmatrix}1&1\\0&1\end{smallmatrix})$ and $x = (\begin{smallmatrix}2&0\\0&1\end{smallmatrix})$. Then $a$ has order $7$, $x$ has order $3$, and $xax^{-1} = (\begin{smallmatrix}1&2\\0&1\end{smallmatrix}) = a^2$.
In particular, $a$ and $x$ do not commute (since $xax^{-1} = a^2$ and $a^2 \not= a$), so $G$ is nonabelian.
A general element of $G$ looks like
$(\begin{smallmatrix}2^i&j\\0&1\end{smallmatrix}) =
(\begin{smallmatrix}1&j\\0&1\end{smallmatrix})
(\begin{smallmatrix}2^i&0\\0&1\end{smallmatrix}) =
(\begin{smallmatrix}1&1\\0&1\end{smallmatrix})^j
(\begin{smallmatrix}2&0\\0&1\end{smallmatrix})^i = a^jx^i.
$ Herstein writes elements of his $G$ in the form $x^ia^j$. As $i$ and $j$ vary, the set of all $a^jx^i$ equals the set of all $x^ia^j$ since every element of $G$ is the inverse of something in $G$ and $(a^jx^i)^{-1} = x^{-i}a^{-j}$.
A similar construction gives you a nonabelian group of order $pq$ where $p$ and $q$ are primes such that $q \equiv 1 \bmod p$. See the top of page 5 here. You are looking at the case $p = 3$ and $q = 7$.
Best Answer
using the notation of your exercise: if $s=0$ then it is obvious that $A: = f^ig^jf^sg^t = f^ig^{j+t}$. Now if $s=1$, in the proof that you posted, the nice formula $gf = fg^{-1}$ is given. Now $A = f^ig^jfg^t = f^i g^{j-1} (gf)g^t = f^ig^{j-1}fg^{t-1}$. Do this again and again: you will end up with $A = f^{i+1}g^{t-j}$. You can give the general formula as $A = f^{i+s}g^{t + (1- 2s)j}$. I hope this answers your question